3.178 \(\int \frac{c+d x+e x^2+f x^3+g x^4}{(a+b x^4)^4} \, dx\)

Optimal. Leaf size=437 \[ -\frac{\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}-\frac{8 a f-x \left (a g+11 b c+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac{x \left (7 (a g+11 b c)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}+\frac{x \left (-a g+b c+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3} \]

[Out]

(x*(b*c - a*g + b*d*x + b*e*x^2 + b*f*x^3))/(12*a*b*(a + b*x^4)^3) + (x*(7*(11*b*c + a*g) + 60*b*d*x + 45*b*e*
x^2))/(384*a^3*b*(a + b*x^4)) - (8*a*f - x*(11*b*c + a*g + 10*b*d*x + 9*b*e*x^2))/(96*a^2*b*(a + b*x^4)^2) + (
5*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(32*a^(7/2)*Sqrt[b]) - ((77*b*c + 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*ArcTan[1 -
(Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(5/4)) + ((77*b*c + 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*ArcTan
[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(5/4)) - ((77*b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*L
og[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/4)*b^(5/4)) + ((77*b*c - 15*Sqrt[a]*
Sqrt[b]*e + 7*a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/4)*b^(5/4))

________________________________________________________________________________________

Rubi [A]  time = 0.531377, antiderivative size = 437, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1858, 1854, 1855, 1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}-\frac{8 a f-x \left (a g+11 b c+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac{x \left (7 (a g+11 b c)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}+\frac{x \left (-a g+b c+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^4)^4,x]

[Out]

(x*(b*c - a*g + b*d*x + b*e*x^2 + b*f*x^3))/(12*a*b*(a + b*x^4)^3) + (x*(7*(11*b*c + a*g) + 60*b*d*x + 45*b*e*
x^2))/(384*a^3*b*(a + b*x^4)) - (8*a*f - x*(11*b*c + a*g + 10*b*d*x + 9*b*e*x^2))/(96*a^2*b*(a + b*x^4)^2) + (
5*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(32*a^(7/2)*Sqrt[b]) - ((77*b*c + 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*ArcTan[1 -
(Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(5/4)) + ((77*b*c + 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*ArcTan
[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(5/4)) - ((77*b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*L
og[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/4)*b^(5/4)) + ((77*b*c - 15*Sqrt[a]*
Sqrt[b]*e + 7*a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/4)*b^(5/4))

Rule 1858

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]] /; G
eQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1854

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[((a*Coeff[Pq, x, q] -
b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q, x])*(a + b*x^n)^(p + 1))/(a*b*n*(p + 1)), x] + Dist[1/(a*n*(p + 1))
, Int[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^(p + 1), x], x] /; q == n - 1] /
; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rule 1876

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(x^ii*(Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2)))/(a + b*x^n), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^4} \, dx &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}-\frac{\int \frac{-11 b c-a g-10 b d x-9 b e x^2-8 b f x^3}{\left (a+b x^4\right )^3} \, dx}{12 a b}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac{\int \frac{-7 (-11 b c-a g)+60 b d x+45 b e x^2}{\left (a+b x^4\right )^2} \, dx}{96 a^2 b}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac{x \left (7 (11 b c+a g)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}-\frac{\int \frac{-21 (11 b c+a g)-120 b d x-45 b e x^2}{a+b x^4} \, dx}{384 a^3 b}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac{x \left (7 (11 b c+a g)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}-\frac{\int \left (-\frac{120 b d x}{a+b x^4}+\frac{-21 (11 b c+a g)-45 b e x^2}{a+b x^4}\right ) \, dx}{384 a^3 b}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac{x \left (7 (11 b c+a g)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}-\frac{\int \frac{-21 (11 b c+a g)-45 b e x^2}{a+b x^4} \, dx}{384 a^3 b}+\frac{(5 d) \int \frac{x}{a+b x^4} \, dx}{16 a^3}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac{x \left (7 (11 b c+a g)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac{(5 d) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,x^2\right )}{32 a^3}+\frac{\left (77 b c+15 \sqrt{a} \sqrt{b} e+7 a g\right ) \int \frac{\sqrt{a} \sqrt{b}+b x^2}{a+b x^4} \, dx}{256 a^{7/2} b^{3/2}}-\frac{\left (15 e-\frac{7 (11 b c+a g)}{\sqrt{a} \sqrt{b}}\right ) \int \frac{\sqrt{a} \sqrt{b}-b x^2}{a+b x^4} \, dx}{256 a^3 b}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac{x \left (7 (11 b c+a g)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}-\frac{\left (77 b c-15 \sqrt{a} \sqrt{b} e+7 a g\right ) \int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx}{512 \sqrt{2} a^{15/4} b^{5/4}}-\frac{\left (77 b c-15 \sqrt{a} \sqrt{b} e+7 a g\right ) \int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx}{512 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\left (77 b c+15 \sqrt{a} \sqrt{b} e+7 a g\right ) \int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{512 a^{7/2} b^{3/2}}+\frac{\left (77 b c+15 \sqrt{a} \sqrt{b} e+7 a g\right ) \int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{512 a^{7/2} b^{3/2}}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac{x \left (7 (11 b c+a g)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}-\frac{\left (77 b c-15 \sqrt{a} \sqrt{b} e+7 a g\right ) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\left (77 b c-15 \sqrt{a} \sqrt{b} e+7 a g\right ) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\left (77 b c+15 \sqrt{a} \sqrt{b} e+7 a g\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}-\frac{\left (77 b c+15 \sqrt{a} \sqrt{b} e+7 a g\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}\\ &=\frac{x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac{x \left (7 (11 b c+a g)+60 b d x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac{8 a f-x \left (11 b c+a g+10 b d x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}-\frac{\left (77 b c+15 \sqrt{a} \sqrt{b} e+7 a g\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\left (77 b c+15 \sqrt{a} \sqrt{b} e+7 a g\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{5/4}}-\frac{\left (77 b c-15 \sqrt{a} \sqrt{b} e+7 a g\right ) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}+\frac{\left (77 b c-15 \sqrt{a} \sqrt{b} e+7 a g\right ) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{5/4}}\\ \end{align*}

Mathematica [A]  time = 0.366398, size = 411, normalized size = 0.94 \[ \frac{-\frac{256 a^{11/4} \sqrt [4]{b} (a (f+g x)-b x (c+x (d+e x)))}{\left (a+b x^4\right )^3}+\frac{32 a^{7/4} \sqrt [4]{b} x (a g+11 b c+b x (10 d+9 e x))}{\left (a+b x^4\right )^2}+\frac{8 a^{3/4} \sqrt [4]{b} x (7 a g+77 b c+15 b x (4 d+3 e x))}{a+b x^4}-6 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (80 \sqrt [4]{a} b^{3/4} d+15 \sqrt{2} \sqrt{a} \sqrt{b} e+7 \sqrt{2} a g+77 \sqrt{2} b c\right )+6 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (-80 \sqrt [4]{a} b^{3/4} d+15 \sqrt{2} \sqrt{a} \sqrt{b} e+7 \sqrt{2} a g+77 \sqrt{2} b c\right )-3 \sqrt{2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )+3 \sqrt{2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-15 \sqrt{a} \sqrt{b} e+7 a g+77 b c\right )}{3072 a^{15/4} b^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^4)^4,x]

[Out]

((8*a^(3/4)*b^(1/4)*x*(77*b*c + 7*a*g + 15*b*x*(4*d + 3*e*x)))/(a + b*x^4) + (32*a^(7/4)*b^(1/4)*x*(11*b*c + a
*g + b*x*(10*d + 9*e*x)))/(a + b*x^4)^2 - (256*a^(11/4)*b^(1/4)*(a*(f + g*x) - b*x*(c + x*(d + e*x))))/(a + b*
x^4)^3 - 6*(77*Sqrt[2]*b*c + 80*a^(1/4)*b^(3/4)*d + 15*Sqrt[2]*Sqrt[a]*Sqrt[b]*e + 7*Sqrt[2]*a*g)*ArcTan[1 - (
Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 6*(77*Sqrt[2]*b*c - 80*a^(1/4)*b^(3/4)*d + 15*Sqrt[2]*Sqrt[a]*Sqrt[b]*e + 7*Sqrt
[2]*a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] - 3*Sqrt[2]*(77*b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*Log[Sqrt[
a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + 3*Sqrt[2]*(77*b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*Log[Sqrt[a]
+ Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(3072*a^(15/4)*b^(5/4))

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Maple [A]  time = 0.014, size = 560, normalized size = 1.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x)

[Out]

(15/128*e/a^3*b^2*x^11+5/32*d/a^3*b^2*x^10+7/384*(a*g+11*b*c)/a^3*b*x^9+21/64/a^2*b*e*x^7+5/12/a^2*d*b*x^6+3/6
4/a^2*(a*g+11*b*c)*x^5+113/384/a*e*x^3+11/32*d/a*x^2-1/128*(7*a*g-51*b*c)/a/b*x-1/12*f/b)/(b*x^4+a)^3+7/512/b/
a^3*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)*g+77/512/a^4*c*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2
)/(1/b*a)^(1/4)*x+1)+7/512/b/a^3*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)*g+77/512/a^4*c*(1/b*a
)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)+7/1024/b/a^3*(1/b*a)^(1/4)*2^(1/2)*ln((x^2+(1/b*a)^(1/4)*x*2
^(1/2)+(1/b*a)^(1/2))/(x^2-(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2)))*g+77/1024/a^4*c*(1/b*a)^(1/4)*2^(1/2)*ln((x
^2+(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2))/(x^2-(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2)))+5/32/a^3*d/(a*b)^(1/2)*
arctan(x^2*(b/a)^(1/2))+15/1024/a^3*e/b/(1/b*a)^(1/4)*2^(1/2)*ln((x^2-(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2))/(
x^2+(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2)))+15/512/a^3*e/b/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*
x+1)+15/512/a^3*e/b/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**4+f*x**3+e*x**2+d*x+c)/(b*x**4+a)**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.09294, size = 629, normalized size = 1.44 \begin{align*} \frac{\sqrt{2}{\left (40 \, \sqrt{2} \sqrt{a b} b^{2} d + 77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac{1}{4}} a b g + 15 \, \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x + \sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{512 \, a^{4} b^{3}} + \frac{\sqrt{2}{\left (40 \, \sqrt{2} \sqrt{a b} b^{2} d + 77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac{1}{4}} a b g + 15 \, \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x - \sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{512 \, a^{4} b^{3}} + \frac{\sqrt{2}{\left (77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac{1}{4}} a b g - 15 \, \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \log \left (x^{2} + \sqrt{2} x \left (\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{b}}\right )}{1024 \, a^{4} b^{3}} - \frac{\sqrt{2}{\left (77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac{1}{4}} a b g - 15 \, \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \log \left (x^{2} - \sqrt{2} x \left (\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{b}}\right )}{1024 \, a^{4} b^{3}} + \frac{45 \, b^{3} x^{11} e + 60 \, b^{3} d x^{10} + 77 \, b^{3} c x^{9} + 7 \, a b^{2} g x^{9} + 126 \, a b^{2} x^{7} e + 160 \, a b^{2} d x^{6} + 198 \, a b^{2} c x^{5} + 18 \, a^{2} b g x^{5} + 113 \, a^{2} b x^{3} e + 132 \, a^{2} b d x^{2} + 153 \, a^{2} b c x - 21 \, a^{3} g x - 32 \, a^{3} f}{384 \,{\left (b x^{4} + a\right )}^{3} a^{3} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x, algorithm="giac")

[Out]

1/512*sqrt(2)*(40*sqrt(2)*sqrt(a*b)*b^2*d + 77*(a*b^3)^(1/4)*b^2*c + 7*(a*b^3)^(1/4)*a*b*g + 15*(a*b^3)^(3/4)*
e)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^3) + 1/512*sqrt(2)*(40*sqrt(2)*sqrt(a*b)
*b^2*d + 77*(a*b^3)^(1/4)*b^2*c + 7*(a*b^3)^(1/4)*a*b*g + 15*(a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2
)*(a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^3) + 1/1024*sqrt(2)*(77*(a*b^3)^(1/4)*b^2*c + 7*(a*b^3)^(1/4)*a*b*g - 15*(a
*b^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4*b^3) - 1/1024*sqrt(2)*(77*(a*b^3)^(1/4)*b^2*c
+ 7*(a*b^3)^(1/4)*a*b*g - 15*(a*b^3)^(3/4)*e)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4*b^3) + 1/384*
(45*b^3*x^11*e + 60*b^3*d*x^10 + 77*b^3*c*x^9 + 7*a*b^2*g*x^9 + 126*a*b^2*x^7*e + 160*a*b^2*d*x^6 + 198*a*b^2*
c*x^5 + 18*a^2*b*g*x^5 + 113*a^2*b*x^3*e + 132*a^2*b*d*x^2 + 153*a^2*b*c*x - 21*a^3*g*x - 32*a^3*f)/((b*x^4 +
a)^3*a^3*b)